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If $f(z)$ is analytic, prove that $$\left(\frac{d^{2}}{dx^{2}}+\frac{d^{2}}{dy^{2}}\right)\left|f\left(z\right)\right|^{2}=4\left|f^{\prime}\left(z\right)\right|^{2}$$

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What is P.T.?... – DonAntonio Feb 15 '13 at 3:00
if you know that $4\frac{\partial}{\partial z}\frac{\partial}{\partial \overline{z}} = \Delta$, then it is easy. – Yimin Feb 15 '13 at 3:18

We will use the following differential operators $$ \frac{\partial}{\partial z}=\frac{1}{2}\left( \frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\qquad \frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left( \frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right) $$ which commute and, for $C^2$ functions, satisfy $$ \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}=4\frac{\partial^2}{\partial \bar{z}\partial z}. $$ Recall Cauchy-Riemann's equation, which characterizes analyticity (holomorphy): $$ \frac{\partial f}{\partial \bar{z}}=0 \quad\Leftrightarrow \quad\frac{\partial \bar{f}}{\partial z}=0. $$

First, we have $$ \frac{\partial}{\partial z}f\bar{f}=\frac{\partial f}{\partial z}\bar{f}+f\frac{\partial \bar{f}}{\partial z}=\frac{\partial f}{\partial z}\bar{f}. $$ Thus $$ \frac{\partial }{\partial \bar{z}}\frac{\partial }{\partial z}f\bar{f}=\frac{\partial^2 f}{\partial\bar{z}\partial z}\bar{f}+\frac{\partial f}{\partial z}\frac{\partial \bar{f}}{\partial \bar{z}}=\frac{\partial}{\partial z}\frac{\partial f}{\partial \bar{z}}\bar{f}+\frac{\partial f}{\partial z}\overline{\frac{\partial f}{\partial z}}=|f'(z)|^2. $$

Since $|f|^2=f\bar{f}$, we now get the formula you wanted.

Let me know if you need more details.

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The question was actually posted by me and I agree with the above solution, but I eventually found the solution in this method and I think this is correct and easily understandable.

$$LHS = \left(\frac{\partial ^{2}}{\partial{x^{2}}}+{\frac{\partial ^{2}}{\partial{y^{2}}}}\right)\left|f\left(z\right)\right|^{2}$$
$$= \frac{\partial ^{2}}{\partial{x^{2}}}{\left(u^2+v^2\right)}+{\frac{\partial ^{2}}{\partial{y^{2}}}}{\left(u^2+v^2\right)}$$
Consider $u^2$ & $v^2$ as functions of functions u & v and apply Chain rule $$= \frac{\partial}{\partial{x}}{\left(2u*u_x+2v*v_x\right)}+{\frac{\partial}{\partial{y}}}{\left(2u*u_y+2v*v_y\right)}$$ Apply the second differentiation using product rule, $$=2\left(uu_{xx}+u_x^2+vv_{xx}+v_x^2+uu_{yy}+u_y^2+vv_{yy}+v_y^2\right)$$ But $u_{xx}+u_{yy}=0$, $v_{xx}+v_{yy}=0$, $u_x=v_y$, & $u_y=-v_x$ $$\therefore LHS=2\left(u_x^2+v_x^2+u_y^2+v_y^2\right)$$ $$LHS=4\left(u_x^2+v_x^2\right)$$

We know that $f^{\prime}\left(z\right)=u_x+iv_x=v_y-iu_y$ Then $$RHS=4\left|f^{\prime}\left(z\right)\right|^2$$ $$=4\left(u_x^2+v_x^2\right)$$ $$=LHS$$ Hence it is proven

Thank you for your help and correct my mistakes in the solution if any.

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